Question

# If a polymer is prepared from 4.90008 g of acrylic acid, and 4.625 mL of 8.0...

If a polymer is prepared from 4.90008 g of acrylic acid, and 4.625 mL of 8.0 M NaOH, what will the mass of the "dry" polymer be after all of the water is removed? Assume all reactions go to completion. Hint: You may want to construct a BCA table to determine the outcome of the neutralization step.

Solution.

The polymerization reaction is The molar mass of acrylic acid is 72.0627 g/mol; the amount of substance is n = m/M = 4.90008/72.0627 = 0.068 moles.

The amount of NaOH is n = cV = 8.0*0.004625 = 0.037 moles. The sodium hydroxide is a limiting reactant.

(The mass of polymer) = (mass of NaOH) + (mass of acrylic acid) - (mass of water);

The mass of NaOH is n*M = 0.037*40 = 1.48 g;

The mass of acrylic acid  is n*M = 0.037*72.0627 = 2.666 g;

The mass of water is n*M = 0.037*18.0153 = 0.6656 g;

The mass of polymer =1.48+2.666-0.6656 = 3.48 g.

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